in a certain city, the monthly water bill amount is normally distributed with mean 25 and...

in a certain city, the monthly water bill amount is normally distributed with mean 25 and standard deviation 15. A sample of 36 bills was selected. What is the probability that the average water bill amount for the sample was between 26.7 and 30? round your answer to 4 decimal places. Do not write as percentage.

Solutions

Expert Solution

Solution :

Given that ,

mean = = 25

standard deviation = = 15

n = 36

= 25

= / $\sqrt$ n= 15/ $\sqrt$ 36=2.5

P(26.7< <30 ) = P[(26.7 - 25) /2.5 < ( - ) / < (30 - 25) /2.5 )]

= P(0.68 < Z <2 )

= P(Z < 2) - P(Z < 0.68)

Using z table

=0.9772 - 0.7517

=0.2255

probability=0.2255

orchestra answered 1 year ago

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